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Proof step 2 from proof prefix-empty-left (
previous step
)
⊦ st0 ⊑ st ↔ dom st0 ⊆ dom st ∧ (∀ n. n ∈ st0 ∧ n ∈ st → st0[n].list ⊑ st[n].list ∧ st0[n].ladrlist ⊑ st[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st[n].list = # st[n].ladrlist)
Lemma
blockstore-prefix